Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 6x}{x + 8} = \dfrac{-19x - 40}{x + 8}$
Answer: Multiply both sides by $x + 8$ $ \dfrac{x^2 - 6x}{x + 8} (x + 8) = \dfrac{-19x - 40}{x + 8} (x + 8)$ $ x^2 - 6x = -19x - 40$ Subtract $-19x - 40$ from both sides: $ x^2 - 6x - (-19x - 40) = -19x - 40 - (-19x - 40)$ $ x^2 - 6x + 19x + 40 = 0$ $ x^2 + 13x + 40 = 0$ Factor the expression: $ (x + 5)(x + 8) = 0$ Therefore $x = -5$ or $x = -8$ At $x = -8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -8$, it is an extraneous solution.